text:
How does engine primer work?
Does anyone have good resources on how piston-engine primers work?
Looking for literature, schematics, or better yet, videos.
I was told that in carburetor engines, it bypasses the carburetor and goes straight into one of the cylinders, but the I was told it goes through a different path in the carburetor.
Thanks in advance!

resource location is off-topic as defined by the help center. You'd better directly ask for what you want to know about primer, precising you better understand with schematics. Good answers are likely to provide links and references for further readings.

Related question: "Why do aircraft piston engines require priming?"  https://aviation.stackexchange.com/questions/78759/why-do-aircraft-piston-engines-require-priming/78775#78775

It's just a syringe pump that draws gasoline from the fuel lines where they pass through the cabin.  The outlet of the primer pump runs though plumbing that becomes a network of very small (usually copper) lines snaking around the engine to one or more cylinders, where it terminates at each cylinder in a small nozzle threaded into the intake port just upstream of the intake valve. This plumbing network is completely separate from the fuel system under the cowling with the gascolator, hoses and carb.
I couldn't find any handy schematics online, so I made a silly sketch.  My own airplane runs primer to 3 cylinders like this.  You can prime all 4 or sometimes you see primer line only to one cylinder (airplanes that don't see much cold weather).

When you stroke the primer, it's like you're a doctor injecting something into your patient, except instead of an unfortunate person, you are pumping fuel into those lines running to the cylinders and spraying a mist of raw fuel into the intake port just up from the intake valve.  The fuel spreads around and settles on the walls of the intake port and duct, and starts to evaporate.
The evaporating fuel provides the additional fuel needed, beyond the carburetor's supply, to make the mixture rich enough for starting and to keep the engine running for a bit while it warms up, and works for a number of seconds after starting before the excess fuel has evaporated (by that time the engine is able to sustain itself on the carb's fuel - aircraft carbs normally run a richer than necessary mixture for the idle circuit, to help them run better when cold).  In non-aviation engines, a choke is used on the carb, but aviation practice traditionally avoids adding potential intake blocking gadgets like choke plates.
The more strokes you pump, the more fuel coats the intake passage.  When it's cold outside, you need more fuel to give a longer evaporation time so that the rich mixture lasts longer, so more strokes are required, although obviously, not too much or you flood it.
The caburetor has an accelerator pump that squirts extra fuel when you pump the throttle as well, but this is to help an already running engine cope with sudden power demands when the throttle is opened quickly.  You can pump the throttle to get extra fuel for starting, they way you used to do in ye olden tymes with carbureted cars, but too much fuel just dribbles down into the airbox when you do that and it creates a fire hazard, so it's not a good practice.
Fuel injected engines are primed in a similar way but they just use the injection system itself to spray fuel in when you turn the boost pump on with the mixture rich.  In those engines you don't stroke a pump, but just put the boost pump on, put the mixture to rich, count a couple of seconds, then pull it back to idle cutoff and hit the starter.

I wasn't able to find any documentation for this, but I can just tell you what I know.
A carburetor mixes fuel and air together into a combustible mixture that's then fed into the cylinders via a series of intake valves. It needs a (small) pool of gas, which is kept topped off by the line from the main gas tank(s). If an engine sits for a while, that tiny little pool of gas will gradually evaporate or leak out. This makes it impossible to start the engine until the reservoir can be refilled.
So, to start the engine, you need to pump enough fuel into that reservoir, so that enough gas will reach the engine for it to run. This is what "priming" the engine actually means. There are several different ways to do this:

On planes with a direct-drive fuel pump, you can continuously crank the engine to pump gas into the carburetor, until enough accumulates for ignition. This procedure usually isn't recommended, because the starter can overheat or drain the battery if you crank it for too long.

On planes with an electric fuel pump, you can run the pump for a few seconds to prime the engine.

Many plane's throttles are designed to pump a small squirt of gas into the carburetor when the throttle is pushed full forward. So, on these planes, you can prime the engine just by moving the throttle back and forth a few times.

Some (mostly older) planes have a separate primer knob. This is just a small, manually operated gas pump that bypasses the mixture control.

Another option is to "choke" the engine. This does exactly what it sounds like, restricts the flow of air into the carburetor so that gas is drawn into it via suction.

If all else fails and you really need to get in the air, you can theoretically open the carburetor and just pour gas into it. I've actually seen this done to an old carbureted car.

This doesn't attempt to answer the question about how primers work.

Priming does not pump fuel into a "reservoir". Your understanding of aircraft fuel systems is incorrect.

Thank you so much!!


text:
Why not always take off with 10 degree flaps in a Cessna 172?
10 degree flaps increase our lift, reduce stall speed, reduce ground roll.
Seems like all advantages with no disadvantages.
So why does the Cessna 172 POH say normal take off is 0-10 degree flaps?
Why not always use 10 degree?

Welcome to aviation.SE! This question might be helpful.

In some headwinds you might take off backwards.

10 degrees flaps is standard in a 177. There are wind situations where less flaps make the plane mor controllable. I don't think I've ever had to. If the elevator trim is set correctly, basically where it was for the prior landing, the plane will lift off with very littl back pressure.  I've only flown a 172 once, for a BFR so I could move my plane on a ferry permit, and I think I took off the same way. May not have used flaps, I don't recall what that plane's check list said.  I do remember that awkward sitting up in a high-back chair feeling to see out the front; the Cardinal is much nicer.

"trim is set correctly, basically where it was for the prior landing" -- don't do that. Take-off checklist specifically requires neutral trim. I've crashed a C172 simulator because of out-of-the-way initial trim made the plane uncontrollable, and I couldn't figure out why until ground.

Flaps out will reduce the ground run, but you're forgetting that they also increase drag. This is why you don't climb all the way to cruise altitude with flap extended. A 172 will climb better without flaps.
With a take-off, you have to consider both the ground run and initial climb. After all, the take-off distance required is defined as the distance required to climb to screen height (50ft in Australia, sorry if the US is different). The aircraft also has to meet a legal minimum climb gradient to that height.
Without looking at the data, it is possible that in certain conditions, taking off with flaps 10 will get you off the ground quicker, but will actually take longer to reach 50ft compared to a takeoff with no flaps. In any case, retracting flaps in a busy time after takeoff is a potential hazard, and in a training aircraft like the 172 simple procedures are best.

Also, in many places the runway is just way longer than a 172 needs, and it doesn't really matter if you take of at 1/4 of the runway or at 1/3 -- but the faster takeoff takes you out of the airport control zone faster.

Using flaps creates drag.  If you have a nice, smooth paved runway, it's more efficient to take off with no flaps.  On a rougher surface like grass or dirt, you have more drag from the wheels, so it's better to get them off the ground sooner.

Also, in gusty crosswinds, the increased flaps & drag make it harder to control the aircraft.

And screen height is 35 ft in the US.

@Kolom I’m not sure what “screen height” is, but the performance charts for my C172 show distance to clear a 50ft obstacle.

The issue is how one wants to optimize their flight operations.  Generally the emphasis is on time to climb, however for some operations, such as out of standing snow, slush, mud and grass, and in the case of seaplanes, glassy water, the preferred optimization may be to minimize ground roll, rolling drag, tire wear or other factors.  I teach both techniques, so that the pilot understands the choices he has, and the ways he can accomplish things.
Concerns about "stalling out" after rotation, are an issue of proficiency.  Correct instruction and recurrent practice help address proficiency.
Probably 2/3 of my flight review students rarely use flaps in a C172 for take off, and generally only when doing soft field or turf takeoffs.  So with those students, we tend to review takeoffs with flaps, and how they may be operationally effective in situations other than just a turf or dirt field.  For those more fluent in flapped takeoffs, there are other things we might spend time on.
From a personal perspective, I probably use 10 deg flaps on a C172 95% of the time, because I like to minimize wear on my tires.

Using flaps is really like saying why not add 4 feet of wing on either side.  Same answer, safer in wind taking off and landing without them.  Tricycle gear also helps by keeping AOA low before rotation.  You don't want to pop up into the air with barely enough speed to fly.  Really the same for landing.  students must realize flaps increase AOA, and will stall that part of the wing at a lower nose AOA.

There’s nothing saying that you can’t. And for a performance takeoff like a short field take off, you will use 10° of flap. It’s just in general conditions combined with the fact of the Cessna 172 only has a takeoff roll around 1000 feet or so at sea level conditions, there’s not much need to use flaps. You can use up to 10° of flaps if desired. But generally most pilots don’t .

With 10 degrees of flaps and the trim wheel in the "take-off position" my C-172 will rotate without any back pressure on the yoke. You have to apply forward pressure to keep the speed above 62 knots. A flight instructor a few years ago during a flight review suggested I always use 10 degrees of flaps during take-off. After using this method for a few months, I concluded that using 10 degrees introduced a new hazard. A stall during take-off was more likely. I went back to 0 degrees flaps.

This answer could benefit from some sources other than your personal experience

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
What are these holes at the rear of 747s?
What are these two holes at the rear of most 747s? I've found plenty of answers for the holes on the front of the wing box, but no answers about these back holes.

What are these elements on the back of a Boeing 737? with a photo of what is behind the hole.

If this question had a picture of a 737, I'd agree that it's a dupe of those questions.  But it's a different type of aircraft with two holes, while the 737 has one.  To close as a dupe is to imply that those 2 on the 747 are the same thing as the 1 on the 737, although without actually SAYING so, nor having a source that TELLS us so.  If the 747 does in fact have two outflow valves (believable; I don't know offhand), I'd think it's reasonable to provide an answer that documents this fact, rather than implying as much without having to support it by 'closing as dupe.'  VTLO.

These are the cabin outflow valves for regulating cabin pressure.

Can you provide a source so that somebody who questions if the 747 does in fact have two outflow valves (while smaller aircraft have just one), has something they can see to verify what you've posted?  I can believe that your answer is probably correct, but documentation would be good.

Unfortunatelly the manual I am refering to is copy right protected but to quote a small section: "Two outflow valves are installed at the rear of the cabin. The valves normally
operate in parallel. Cabin altitude and full ventilation rates can be maintained by
either valve." (Boeing 747 Operations Manual)

Here is the overhead panel section containing the valves controls. There are two sets of controls below the yaw damper switches. Each set corresponds to a valve.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
How is a plane with only ailerons and jet (no props) controlled?
I saw this video and was wondering how he is controlling the plane:

It looks like he only has ailerons. It does look like he is shifting his body weight. Is that the answer? If so, that's pretty surprising! But I guess it's like a hang glider?
Does anyone know how you should shift your body to increase/decrease pitch, and increase/decrease yaw?

The whole device shown in the video is quite scary. What protection exists for the pilot when landing in a field in an emergency where the likelihood of the plane toppling over is quite real? And how did they pan and zoom the installed cameras? Remote control?

@PeterKämpf - What? I take it that you don’t think the Walmart bicycle helmet is enough?

As Peter Kampf has said, the aileron control surfaces also act as elevators. I wonder if the body movement is negligible or necessary. Does the weight shifting aid in countering adverse yaw? Or, is there a greater degree of deflection when the aileron is up versus when it is down? Is the difference adjustable in flight? If not, does that affect the aircraft’s crosswind capability? After all, there are aircraft like the Eurocoupe and flying wings that fly without rudders.

"Livin' on a prayer' comes to mind.

@PeterKämpf - most likely they used 360 degree cameras, and the pan/tilt were done during video editing.

@DeanF. I guess weight shifting is needed for directional control on the ground. During take-off the thing was almost uncontrollable and also the landing was not straight but diagonal across the runway. Regarding differential ailerons: That would couple roll and pitch and should be done in moderation only.

What a death trap.  I call it the "Parapleg-O-Craft".

Propellers, in prop driven aircraft, don't offer any flight control other than thrust, so the fact that this is a jet driven craft vs a prop driven craft makes no difference.

Limited market for this craft as young Anakin Skywalker is the only human who can fly it.

@quietflyer According to this interview (Warning: Japanese), the harness does control roll through control cables to the ailerons, but pitch is entirely done through shifting the CG. The throttle is in a handle on the pilot's right.

@RobinBennett possibly - when you can see one from another they don't look like any 360°cameras I can find, and they don't show enough distortion or evidence of flattening unless they've got pixels to spare.  I'm pretty certain the one on the wing is a normal GoPro, but that doesn't pan, unlike the ones by the pilot's hip and on the nosewheel strut.  The one on the nosewheel strut must be either 360° or panned though, from the first few seconds

@J... A similar question said that you can use the torque differential between propellers to control a plane, not unlike how a drone uses torque differential to yaw. Actually the plane in the other question looks almost the same as the plane in my question, with the only difference being a jet rather than props: https://aviation.stackexchange.com/questions/46311/would-an-aircraft-with-no-ailerons-no-elevators-and-no-rudders-be-safely-flyabl?rq=1

@Croolsby You can, but typically only in an emergency.  The aircraft is never designed to use twin engines for manoeuvring.

The ailerons in most flying wings can also move together up and down, then working as an elevator. This is called an "elevon", combining both words as the device is combining both functions. Since flying wings mostly use reflex airfoils, loads on the rear part of the wing tend to be small and the force to keep elevons at their right deflection is small, too. In case of the SB-13, they are pre-loaded by a small, adjustable spring but otherwise free-floating.
In this case, however, it could be that they only work as ailerons.
When used for pitch control, elevons must move up and down together on both sides. This means they will be freely floating and must be held in position by some force. It would be easier to cross-link them so they only can move against each other, and then they only work as ailerons.
The airplane, called OpenSky M-02, is the brainchild of Tokyo University of the Art professor Kazuhiko Hachiya and has been built by Aircraft Olympos of Japan. OpenSky M-01 was a lighter, unpowered version.
As Louise Hose writes on Kitplanes:

Hachiya designed the project and is the only pilot to fly the resulting “Open Sky M-02J”. He started with a hand-launched, scale model to test the airfoil characteristics. Once he was satisfied with the basic design, he partnered with aeronautical engineer Tota Ueno of Olympos Company in Tokyo to scale up to a glider capable of carrying one person…Hachiya himself. The glider, first flown in 2006, was launched from the ground by teams of people pulling a bungee cord. Once satisfied with its flight characteristic, Hachiya and Ueno worked to find the “right” engine to add as the powerplant. Between 2010 and 2013, they were doing high-speed taxi tests and working through issues…particularly powerplant and regulatory issues. They ultimately decided to use the NIKE jet engine made by AMT Netherlands which produces 176 pounds force. The jet was designed and mostly used in target drones.

At least the model couldn't be steered by weight shifting, and looking at the harness of the pilot, his capacity for weight shifting is rather limited. My impression is that weight shifting is primarily used for ground control. If you watch the YouTube movie, it becomes apparent that he is barely capable of controlling the direction on the ground, and has to lean to the left on take-off to prevent it to escape the runway. Also, his movements are more sideways than back-and-forth. The control surfaces hardly move, but show linkages on the bottom side, so they are moveable. If you look closely, you can see pulleys on the left and right of the pillow Hachiya-san is lying on, and as @quietflyer says, they should control the elevons, but he adds his weight to help controlling both pitch and roll.
Louise Hose of Kitplanes again:

To date, Hachiya has made about 150 “jump” flights and 70 circling flights. He has flown it up to nearly 500’ agl but believes it can fly much higher. (Airspace restrictions in Japan have prevented him from trying to fly higher.)

The ailerons are a little too close to the pitch axis to provide meaningful pitch control. It's pitched with weight shift.

@HiddenWindshield If you really believe that, then study flying wings like the AV-36 more closely. With the little pitch inertia and damping of a straight-wing flying wing, all it needs is a change in the pitching moment coefficient of the airfoil.

Yes, elevons in this configuration could provide some pitch control, but not enough to be practical. The AV-36 has a much longer chord, moving the elevator further back from the CG. And Hachiya himself has said (note: you'll need a translator service unless you read Japanese) that pitch control is done by moving the center of gravity.

@HiddenWindshield: The AV-36 has a much higher aspect ratio than the M-02, so I wonder how that can be combined with a longer chord. But yes, not much movement is needed for pitch control when pitch inertia and damping are small.

Well, I'm no physics expert, so I'm going to stop making claims that I can't back up. But every article I can find on it says the control surfaces are just ailerons, not elevons, regardless of if elevons would work in this situation or not.

@HiddenWindshield: I have now studied the video a bit more and it is obvious that the pillow the pilot is lying on has pulleys connected left and right. By shifting his body around, he will be able to deflect the elevons both symmetrically (for pitch) and against each other (for roll). Weight helps, but is not all. The elevons have their fair share as well. Also, being mounted on the inner wing segment only, those surfaces make rather inefficient ailerons.

@HiddenWindshield Thinking a bit more about this, the pulleys are much better for roll control than pitch. Also, for pitch the control surfaces must be free-floating while to serve as ailerons only they can be linked so they can carry a load without floating up or down. I guess you are right, in that case they really are only ailerons.

2:50 in the video (https://youtu.be/rFeXtspp8OA?t=170) is a decent side view of exercising some pitch control (near the ground so it provides a good reference).  We definitely see the pilot shift his body forward and back.  It's hard to see the far aileron (elevon?).  I can't tell whether it's moving as much (in the opposite direction) as the left aileron goes up and then down.

This is the M-02J, a jet plane modeled after the Mehve "glider" (actually a futuristic antigravity plane) from "Nausicaä of the Valley of the Wind". And, yes, the pilot controls pitch primarily by shifting his weight forward and back, while the ailerons are only used for roll control.
You can control any airplane through weight shift, that's not limited to hang gliders. It's just not practical for larger aircraft, but for little one-man planes, it's perfectly reasonable.
To answer the last question, it works just like a hang glider. The plane's pitch and roll can be controlled by shifting your weight in the direction you want to turn. Yaw isn't controlled, you just have to rely on the natural aerodynamics of the aircraft to keep your nose pointed in the direction of flight.

Case in point. If you are flying right seat in a small, GA aircraft like the Piper Cherokee/Warrior/Archer variety, you can make a trimmed out straight and level aircraft climb by reclining your seatback.

Kudos for correctly identifying the origins of the plane!

The Kitplanes article suggests that the pilot's left-right (and fore-and-aft?) body movements shift the small platform or cradle that he is lying on, which pull on the control cables that actuate the ailerons or elevons, whichever they are.  No joysticks appear to be involved. We lack an authoritative answer as to whether pitch control is primarily/ partially via elevons or solely weight-shift, but the former seems more likely. Even if pitch control is mainly through elevons, the effect of even a slight fore-and-aft shifting of pilot's shifting body weight may not be negligible on a tailless aircraft with little or no wing sweep.
Note that the system is reminiscent of the hip cradle used in the Wright brothers' earliest aircraft, at least as far as roll control goes.
Yaw control is a non-issue with many well-designed aircraft.  For example, very few hang gliders have had yaw control surfaces.

Excellent observation. If you look closely, there are pulleys connected to the pillow on which Hachiya-san is lying.

@PeterKämpf and quiet flyer: I'm pretty sure they're elevons.  I think I've reverse-engineered a plausible design for how those ropes are hooked up to them, which I posted as an answer.  It might not be correct, but I think it's plausible.

I think I see how the elevon controls are designed.  The pulleys by the pilot's waist look independent, except for being connected to the sides of the "pillow" the pilot rests his torso on.  So that creates linkage between controls for side-to-side or forward-backward movement.
The pilot moving his body (and thus the pillow) left or right will pull on one and create slack in the other, creating opposite changes in elevon position.  This gives roll control.
Near the end of the video, after landing (e.g. at 5:00) we see a good angle of the "pillow" positioned so the cord attachment points are as close as they can get to the pulleys.  So the pull on each cord is at a minimum.  This is presumably downward deflection of both elevons since the pilot moved to this position after touchdown.  Moving left or right could probably lower one a bit further (while raising the other).  (In that same shot, we can see the elevons in frame as well, and it looks like they're not far down, but this is a bad angle.  Maybe they simply don't go very far down, or there's some kind of internal linkage, or I'm missing some other control input.)
But in flight, I think the pilot generally keeps his body position farther back, so both ropes are pulled out some.  This pulls the elevons up to a neutral or higher position, leaving room for pitch up or down by moving the pillow back or forward, respectively.  (Along with the pilot's weight, which also provides some pitch control directly, apart from moving the elevons).
In this position, moving left and right will still slacken one and tighten the other, creating the opposite movements you want for roll control.
During landing (4:22), we have a camera angle from above/behind the pilot that shows both ropes and pulleys while he's moving, controlling pitch while maintaining some right-roll.  Note how he pushes backward (tensioning the ropes) to flare a bit just before touchdown.
To make roll control work in the correct orientation, I think the right rope must be connected to the left elevon, and vice versa.  Because moving your body (including bodyweight) right slackens the right rope.  But to roll right, you need to deflect the right elevon up.  If it was right rope => right elevon, either roll or pitch control would have to work against the pilot's weight shifts, depending on whether pulling deflected upward or downward.  That would be terrible and unintuitive, and we can see from the video that he's moving and/or leaning right to roll right, and back to pitch up.
This is my explanation of how I think it's designed, not based on any statements from the designer.  (But thanks to the other answers for their observations that shaped my thinking, especially @quietflyer for spotting the pulleys.)  I think it's a plausible design for the controls, compatible with sanity and all the observations I could make from the video, but possibly not the actual design.
Certainly body-weight shifts are part of the controls; as @quietflyer says we can't be sure how much of the control authority comes from body-weight vs. elevons.  But it looks like there is good control of the elevons for both pitch and roll.

2:50 in the video is a decent side view of exercising some pitch control (near the ground so it provides a good reference).  We definitely see the pilot shift his body forward and back.
It's hard to see the far elevon.  I can't tell whether it's moving as much (in the opposite direction) as the left aileron goes up and then down.  If not, that would prove that these are elevons which can also do pitch control, not just linked ailerons.  I think that's the case from what would make sense for the rope/pulley design, but I'm not 100% sure.

i agree on the feasibility of this arrangement but wonder how much load is carried by the elevons in flight and whether that is compensated by a spring. In that case the pilots works against the spring, which is tolerable for short flights. When Lippisch designed his first flying wings, the same happened (without a spring, so the elevons were only held by pulling on the stick) and only after a 3-hour flight to Berlin with an early motorized version did the pilot notice that he had to pull on the stick all the time.

I think this answer is correct; looking closely at the cords and pulleys it is difficult to imagine how fore-and-aft motions of the cradle would not cause same-direction up or down deflection of the control surfaces, so I think they are indeed elevons. If you look at the video around 2:50 there is little evidence of any rolling motion so I think we are seeing the pilot shift his position strictly in the fore-and-aft direction at this moment in the video.

Truth is my answer was based more on the fact that the Kitplanes article referenced the early Wright aircraft's hip-cradle system, than a really close look at what was visible in the video; I think this answer has filled in the missing pieces as to what must be going on.


text:
What thrust is used for thrust to weight ratio?
Performing some preliminary calculations for an aircraft design I have and was wondering what thrust I should use to calculate thrust to weight? The thrust calculations I have, vary due to speed (since in general force is a change in linear momentum) and so the faster the aircraft is traveling, the lower the thrust I have is. Should I use the initial thrust when the aircraft is initially accelerated, an average thrust of some sort or a different thrust estimate altogether?
These are two different equations I have used to estimate thrust (For propellers)

Well, it depends on what you want to know. If you are interested in thrust-to-weight ratio in certain conditions, you should use thrust (and weight) in those conditions.
In terms of standard performance specifications, the maximum static thrust (zero speed, zero altitude, ISA conditions) is normally used. This is often the only thrust point specified anyway. The weight is often less certain; it would be fair to use MTOW for this figure, but some "practical" weight/configuration is often used instead, because it gives better (and possibly "more relevant") result. Technically, the conditions should be explicitly specified.
If you are interested which thrust formula is more correct, this is a separate question.


text:
What are the best methods for calculating lift during different stages of flight?
Supposing I have the all dimensions of an RC airplane: wingspan, NACA airfoil, aspect ratio, fuselage length, etc., what would be the best methods of calculating the expected lift the aircraft should feel during takeoff, climb, and cruising stages? The conditions are assumed to be that of sea level.
I also have estimates of thrust produced by the propellers on the airplane, center of gravity location, stall speed and lift off speed

As we have learned here at Avation Stack Exchange through many writings: thrust or power variation is critical in take off, climb, and cruise flight.
In straight line steady state flight (at the same speed) lift only equals weight in level flight, in climb or descent it is equal to the cosine climb or descent angle x weight.  In a climb the engine must provide the remainder of vertical lift, in a descent drag reduces vertical lift requirement.
Plugging in some numbers for cosine, we see lift is close to weight until climb or descent angles exceed around 20 degrees. (Cosine 10 degrees = 0.98 Cosine 20 degrees = 0.94)
Lift may exceed weight during maneuvering, known as "G's".
This involves curving flight.  Every plane has a G limit, even models.  But in straight line flight, under most conditions, lift produced is fairly close to the weight of the plane, and is governed by airfoil type, wing area, air density, speed, and angle of attack.


text:
Can I Return Back to the Destination Airport after I Diverted to the Destination Alternate?
Can I Return Back to the Destination Airport after I Diverted to the Destination Alternate (while enroute to the destination alternate)?
I highly appreciate an answer with reference please.

Can you please explain why you even need to ask this question? Why do you think it would not be allowed or would cause a problem?

There are no regulations in the US that state you must land at a particular airport. It will be even more difficult to find one that says you can return to your original destination after announcing or performing a diversion. There are only regulations stating that you have to list a destination airport on an IFR flight plan. And, that there are certain situations when you must list one or more alternate destinations on an IFR flight plan.
VFR flight plans are not required in the US. Even if you file a VFR flight plan, ATC will not be made aware of your flight plan unless it is not closed prior to half an hour after your ETA. In which case, only VFR aircraft on VFR Flight Following will be tracked by ATC. The pilot can cancel a flight plan (IFR or VFR) or flight following at any time that the pilot is in VMC, entering Class G airspace or landed. Otherwise, you must fly your filed IFR flight plan all the way to your clearance limit or your alternate (once you have informed ATC of your diversion). But, you may also amend your flight plan or Flight Following at any time.
Case in point. I have diverted to an airport just to take an unplanned bathroom break. Before diverting, I informed  ATC of the diversion as well as requesting to keep my flight plan and transponder (squawk) code. When returning to controlled airspace, I was able to continue onto my original destination.
Another example of continuing on to your original airport after diverting is when the diversion airport is en route to your original airport. Having an alternate airport planned that is halfway to your original airport gives the pilot a convenient out or Plan B if the weather ahead starts turning hazardous. If the weather improves before you reach your alternate, you can continue onto your original (after informing ATC if on an IFR flight plan or Flight Following).

To be clear: the only way ATC will "become aware" of your filed VFR flight plan is if Flight Service calls us to ask if we happen to know where you are. We don't have any way of looking at it without calling FSS ourselves. You can cancel an IFR flight plan or VFR flight following with ATC directly; to close a VFR flight plan you need to call FSS.

No, there's no regulation saying you can't return to the original aiport from your diversion airport because it would make no sense. There's no reference to give on this because there's nothing in the regulations that is specific to this situation. There's no reason to prevent you from going back to a primary, and every reason to allow it as the conditions that make you go to an alternative may change and make the original more preferable for safety or other reasons. A couple of examples:

Changing weather: the primary may be below minimums, so you divert. Conditions change at the primary, making it a safer choice
Temporary closure: you arrive at the primary to find it's been temporarily closed due to an incident, objects on the runway or something like that. You divert, then find out your primary is open again. Going back to the primary is convenient for you and your passengers

The limiting factor is fuel, if you have enough fuel to get to the alternative and back with plenty of contingency then by all means do it if the situation warrants it.


text:
Is it better to reduce drag or increase lift?
Essentially my question is, in general is it better to have a lower drag or a larger lift. I know lift to drag ratios are used as an indicator but I think these don't give the best picture. For example, I'm running a 2D airfoil analysis and for 2 cases the CL/CD changes varies between 80 and 160 (obviously doesn't account for 3D drag). So you'd expect the best option as CL/CD as 160.
However the CL/CD only looks at the % change. So if the % increase in drag is say 70% and for lift it's 10% CL/CD will be larger for the increased drag but the amount of increase in lift force greatly outweights that of the drag (an increase of 175N of lift vs reduction of 14N of drag).
I guess it would depend on the application as well.
Any help would be greatly appreciated, thanks in advance!

Your lift will be prescribed and cannot be freely chosen. So what you need to do is to minimize the drag that goes with this given lift.
In straight flight lift needs to equal weight. In turns, the centripetal force is added to this. At high speed a small angle of attack will produce that lift while at low speed the angle of attack needs to be high. In most cases, takeoff and landing requirements will determine minimum wing size and the maximum lift coefficient of the airplane. In order to not drag around massively oversized wings, most airplanes use flaps to have a decent maximum lift coefficient for takeoff and landing while minimizing their wetted surface at high speed in order to keep drag down.

Thank you, this makes sense. I had forgotten that the lift is pre-determined by the straight and level flight condition so drag is the factor to change. Thanks!


text:
Is it possible to file for a flight level ABOVE an Airway altitude limitation (for high en route airways)?
I'm just curious if it's ever possible to file for, and/or fly, an altitude that is above an airway restriction. Say if the airway has an altitude restriction of FL410 or less. Is it ever possible to fly above that (with a capable plane, of course)? Thanks!

Yes, if as you say your aircraft can do that. The rules change somewhat as to separation, as above FL 450 navigation can be done point to point instead of by navaids. FL 600 or above is usually reserved for military and special purpose use.

Can you cite a FAR or other source saying that airspace above FL600 is "reserved"? It's Class E airspace, and to my knowledge there's no limitation on filing for it or flying in it besides the physical limitations of your aircraft.

FAR §71.71 cites above FL600 as Class E airspace. And I was just curious as to why some airways even have a top altitude restriction. Radar coverage limitations?

@randomhead https://www.faa.gov/air_traffic/publications/atpubs/atc_html/chap9_section_2.html

@JohnWard I can only surmise there is an interest in keeping separation between garden variety jet aircraft, some of the more advanced ones that can fly higher, and the rarified ones that can operate above FL 600.

@Juan nothing there says or (as I read it) even implies that civilian operating above FL600 are prohibited.

@randomhead If you read my comment again, carefully, you will see that I never said that.

@Juan So you are saying that above a specific airway’s top altitude, they aren’t used, and instead point-to-point navigation is used? Can the same points as the airway be used, by coincidence?

Juan, you said the airspace above FL600 is normally "reserved" for military or special use. I agree that the .65 has special provisions for military and/or supersonic flight there, but I'm not seeing where it is "reserved" for that purpose

@JohnWard Ask when you file the flight plan.

@Juan Okay, thanks for your help. This is obviously a hypothetical, as I don't own or plan to rent a plane capable of those altitudes. I do attempt to be realistic when flying my flight sim however, which is why I asked this question. Again, thanks for your input.


text:
Why is the definition of $V_{S1}$ vague wrt configuration?
The definition of $V_{S1}$ is

the stalling speed or the minimum steady flight speed obtained in a specific configuration

from 14 CFR §1.2 Abbreviations and symbols.
Why is this vague in terms of the configuration?
My understanding is that $V_{S1}$ is in terms of flaps retracted but this can't be gleaned from the definition.

Typically, such wording means that the "specific configuration" must be stipulated when the designation is actually used. That is, it may depend on the context.
Indeed, in the body of the document we see for "Light-sport aircraft" (emphasis mine):

A maximum stalling speed or minimum steady flight speed without the use of lift-enhancing devices ($V_{S1}$) of not more than 45 knots CAS at the aircraft's maximum certificated takeoff weight and most critical center of gravity.


text:
How to find capture area of an engine inlet?
I tried to relate intake pressure recovery in terms which do not involve efficiency of an inlet and capture area, i was able to formulate pressure recovery without intake efficiency but now i need to find a way to formulate capture area of inlet without without use of intake efficiency and pressure recovery is it possible ? if yes please share.

It is not so hard, actually.
You need to know the speeds ahead and inside the intake. You can deduce their ratio from the pressure recovery factor, and normally the Mach number inside the intake is between 0.4 and 0.5, so this, combined with your design flight speed, is a good starting point.
In supersonic flow the air is rammed into the intake face, producing a shock. This will be a straight shock in case of a pitot intake or an oblique shock in case of an intake ramp. Use the appropriate shock tables to find the corresponding speed and density ratios.
For subsonic flow you can use a Gedankenexperiment: Your intake is flying through a densely packed array of stream tubes. Some of them will end up inside the intake, others will be pushed aside as they come close to the intake lip and will flow around it. The speeds in those tubes varies between flight speed well ahead of the intake and your eventual intake speed inside the intake. In between, things change continuously in subsonic flow or via shocks in supersonic flow. Now assume that all the air which passes through the intake will have to pass through their corresponding stream tubes, both ahead and inside the intake. No air is allowed to cross between tubes. If the air inside the intake flows faster than flight speed, the flow in the tubes will be accelerated as they approach the engine face, so the intake has to ingest a lot of tubes to keep mass flow the same. Conversely, if flight speed is higher than intake speed, some of the tubes will be pushed aside to get a match between the mass flows ahead and inside the intake. Don't forget to adjust stream tube area for density changes: The change in stream tube cross section has to be inversely proportional to the change in air density as the air inside the steam tube changes speed, like this:
$$\frac{\delta\rho}{\rho}=-Ma^2\cdot\frac{\delta v}{v}$$

i appreciate your effort in answering the question.thank you. But you told me to deduce the ratio of the speeds using the pressure recovery factor right?  but i wont be knowing the pressure recovery factor . the whole effort is to roughly calculate the pressure recovery Pi which do does not have efficiency. is it possible to calculate the capture area using continuity equation?

@vbks Yes, the continuity equation is the mathematical expression of the thought experiment outlined above. For the pressure recovery you start with lossless compression and then apply an appropriate reduction. Well-designed intakes have efficiencies of 97-98%, so the loss is small. And you need to know either the speeds or the pressures, or there is no basis for a calculation.


text:
Is it possible for an indirect (Russian) attitude indicator to work in 3D without tumbling?
For a reference of direct and indirect attitude indicator, see How is a confusion possible between Western and Russian Attitude Indicators?, and https://www.skybrary.aero/index.php/Western_and_Russian_Artificial_Horizon_Formats.
Direct (Western) attitude indicators can be made to work without tumbling, or even made into a "Navball" that also displays directional information:

See the attitude indicator at the center, Source, and Source.
Or alternatively, look at Kerbal Space Program's navball.
Is the same thing possible with the indirect style indicator?
If one plays DCS, he or she should know that this style of indicator would always tumble when maneuvering pass 90 degrees above horizon, even though the gyro has no gimbal limit.
This is understandable, because otherwise the indicator must display the horizon upside down... or am I missing something?
Is there another way I missed which can get around this issue, or is indirect attitude indicators not capable of representing full 3D orientation, like a navball can?

Yes. It's actually easier.
Just consider the fact that an indirect indicator consists of two separate moving parts with one degree of freedom each, rather than one with two (or even three for navball) DoF. Instead of a ball, the pitch indicator can be a ring with continuous rotation, like so:

Note that most modern attitude indicators have a remote (separate) sensor, with the indicator driven mechanically by little servos (like the pictured AGD-1) or, of course, rendered digitally. This makes things even easier: the gimbal lock is a separate problem of the sensor (avoided with a tracking mechanism, pictured on the left here), while indication can be made however we want. You can easily have continuous pitch as well as continuous roll.
However, this is usually not done (on either type of indication), and this is a matter of convention rather than technical limitations. If we allow unrestrained angles, attitude becomes ambiguous: pitch ±180° with zero roll is the same as pitch 0 with ±180° roll. For this reason, the convention is that pitch is limited to ±90°, so only the second of these indications is valid. The necessary consequence is that roll "tumbles" when going past ±90° pitch. At the same time, the pitch indicator just inverts direction (and so mechanically it can be a semi-circle).
As a result, you will consistently see 180° roll, zero pitch no matter how you arrive to it: by a semi-loop or by rolling. This is reasonable given that continuous rolling is much more common.

I see. So, does this mean indirect indicators are not good for, say, space flight?

@Carl, "good" is what is most clear for the pilot. This is not a trivial matter, even for pilots themselves (objective tests may show different results to what pilots believe intuitively). In fact, indirect indication was shown to be significantly better (in terms of error rate), esp. in unusual attitudes. So I'm not sure navbal is better for anything. In aviation, splitting pitch/roll (AI) and yaw (HSI/compass) makes perfect sense: one is "flying", the other is "navigating". Similarly, it can be argued that pitch and roll control are substantially different...

...for the pilot and should be treated (and thus indicated) differently/separately. But in space, the whole mental picture may be different, and the indication should reflect that.

I was thinking that navball can represent an inertial reference frame, while an indirect style assumes that there is a "ground plane" to reference from, and thus is not useful in space. Obviously, I agree that this is true for aviation.


text:
Why don't airliners use light strips on physical controls (yoke/stick, throttle and trim) to indicate who has control - pilot or flight computer?
I've watched a lot of Air Disasters episodes about crashes where pilots thought auto-pilot, auto-throttle, auto-trim were "in control" but they weren't (or vice-versa) due to some innocuous pilot action - like bumping the go around setting. See: https://en.wikipedia.org/wiki/China_Airlines_Flight_140
From a Human Factors engineering perspective, it seems that it should always be obvious who has control of the various flight controls. One simple option is to put light strips on each physical control (yoke/stick, throttle, trim) to indicate who "has control" - the pilot or the flight computer.  If the throttle levers light strip is green, auto-throttle has control.  If the light is off, the pilot has control of the throttle. This may seem a bit simplistic for normal aircraft operations but in times of high stress such a simple system might make a difference - preventing a crash.  Lights like this would of course need to adjust to the ambient light so they're not glaring during night yet visible in bright daylight.

(Re: China_Airlines_140) the copilot was manually flying the ILS when he inadvertently triggered the Go-lever, initiating a manual FD guided Go-around with auto throttle at (G/A)THR. The autopilot, which was OFF, was subsequently engaged by one of the pilots, probably the copilot who as PF, at some point reasoned that engaging the Autopilot would probably result in the safest flight path. This clashed terribly with the Captain's prevailing intent of continuing the approach which the copilot was half-heartedly carrying out. I just think these specifics shine some light on the Q at hand.

My personal opinion is that there's sufficient Autopilot status annunciation in the jetliner cockpits of today. It is effective SOP choices (adhere to "Elements of a Stabilisd Approach") and training emphasis that'll solve the problem. Also the acceptance of 'startle factor' being a major player in these situations.

They do.
On the 737 there is a light in the button for each autopilot and next to the autothrottle switch. As well as annunciation in green text at the top-center of the Primary Flight Display (PFD).
I'll see if I can find, or take, a photo (that isn't copyrighted) to illustrate this.
The indications are there; the challenge can be getting thru the information overload so that the pilots realize what the aircraft is telling them.

Thanks. I've seen auto-pilot light indicators on the flight display. And the info overload you mention - lots of indicators on that display - was why I was thinking that if you put a light strip right on the yoke/stick (maybe it run around the circumference or maybe just on top), that's a more obvious indication of who's in control.  Then you could look at the PFD for details.  Just a thought.

@Keith Hill: Putting the indicators on the actual controls might cause problems for those of us who have non-transparent hands :-)  And would subject them to wear &c that might cause them to fail.


text:
How safe are ATR 72/42 planes?
I'm flying end of December and they seem to have issues with icing? Is this something I should be worried about? Thanks.

If they weren't safe they wouldn't be flying. "Safety first" is not just a catch phrase in aviation, it is how the entire aviation sector operates.

"Safety is our goal!  But let's be realistic."

The ATR icing problem was 25 years ago.  I'm sure you'll make it safely by this point.

If you're on a US commercial airline, then you have less chance of dying on a flight than you do doing anything else you can imagine, including sleeping in your bed at night. (A 75-year lifespan spends about 750,000 man-hours in bed, with one death in those hours.  The deaths-per-flight-hour rate for US airlines is between 1/10th and 1/100th of that.)
If the airplanes couldn't meet the safety standards of the FAA, they wouldn't be flying.

I am flying in Europe, also why did boeing 737 MAX meet the FAA and EASA standards then?

In Europe you can also be assured that the aircraft flying commercial routes are considered and proven safe. Icing is a well known fact of aviation and pilots and ground crews know how to manage it.

That's what you might call a Black Swan event with unique circumstances and decisions resulting in something tragic.  Doesn't change the overall risk level in the big picture.  There hasn't been a crash of a Heavy in the US in nearly 20 years.  There used to be one every couple of years a couple generations ago.  The ATR's had a problem with aileron reversal with ice on the leading edge after one rolled over and crashed many years ago.  It was fixed.  As the old saw goes, the drive to the airport is far more dangerous from a pure statistical risk perspective.

@CupOfGreenTea  The Max, even with the bad software, in the hands of competent pilots, was fine.  The problem with both Max crashes was that the pilots in both crews fell tragically short of the standards of experience & hand-flying competence to which American, and many European, crews are held to.  Depending on what airline you're on in Europe, you may be fine -- Lufthansa & British Airways are excellent.  Some from the Former Soviet Union, perhaps not as much.

I'm flying end of December and they seem to have issues with icing?

Remember that every single commercial aircraft flies every single day. Probably multiple flights per day. For most of these flights you're not on board, and don't care - but they still happen.
If an airliner crashes, it's headline news everywhere in the world. Thus we hear about pretty much every aircraft accident, no matter where it happens. I read about such things a couple of times per year - often a regional or commuter air line in some remote country. I've not read about major crashes in Europe this year.
Airline crashes simply doesn't happen, at least not in the western world. Whenever they happen they're basically a combination of multiple bad things at once, and more or less impossible to predict.


text:
Is it possible that as the Reynolds number increases, a laminar bubble increases in extension?
I'm studying the Benedek 10355 airfoil (Plot), which should be a profile used in air-controlled models.
While I was carrying out a boundary layer analysis, I realized the particularity that I have placed in the title.
By making an incompressible analysis at Re = 500,000 and alpha = 2 deg, the profile has a laminar bubble on the upper surface (H index which diverges from 2.5 to 4 and then falls to 1.4, negative friction coefficient and plateau of the pressure coefficient) and then it also looks like a bubble on the lower surface.
In this last case, the H index starts from 2.5 reaches about 3.5 then drops to values of 2.5, a sign that the boundary layer separates but there is no turbulent transition. The friction coefficient assumes slightly negative values.
For Re = 1e6, alpha = 2 deg, the extension of the bubble on the upper surface decreases, while the bubble on the lower surface increases; H starts at 2.5 goes to 4 then drops back to 2.5 (so once again there is no transition on the belly).
A Re = 8e6, no bubbles, either on the upper or on the lower surface.
What is the cause of this abnormal behavior on the lower surface? It happens at every angle of attack, at every Reynolds.
Thanks to those who will be able to help me!

The laminar "bubble" on the bottom is caused by the wavy contour. There is a pressure rise at about 25% on the lower side which causes the boundary layer to thicken, but the subsequent flat pressure profile keeps the boundary layer laminar until it separates at the trailing edge. Here is the plot from XFOIL and I placed a green circle over the spot:

With a higher Reynolds number the flow is less stable, the boundary layer is thinner and the waviness makes itself felt more than at lower Reynolds numbers. That explains the larger "bubble" at Re = 1,000,000. With Re = 8,000,000 the boundary layer transits to turbulent flow ahead of the disturbance and no "bubble" can be observed.
I guess you play around with XFOIL. Go to MDES and PLOT the Qspec, then MODI the contour so it becomes more smooth. The result below (first pressure, then H) should prove my point:

Wow! Thank you so much, your answer  was very helpful! I also noticed that if I play around with the Ncrit, the "bubble" disappears for every Reynolds and every alpha

Both your images show the same Reynolds (0.5*10^6) did you mean to post a different image?

@Federico Yes, that is on purpose. The second one uses a smoothed contour. Compare the incompressible pressure coefficient (dashed line) for the lower side.

ah, I understood that the Re was supposed to be different too, sorry.

I would like to ask you a question; the pressure peak occurs in the non-viscous or viscous case? Because from the Cp graph I see an edge for the dotted line, in the first case.

@leo95nf There is a pressure (better: suction) peak for both, but in the viscous case it is smoothed a bit by viscous effects. That the dotted line shows a sharp peak is not relevant, a suction peak is a minimum along the c$_p$ curve.

Thank you so much!


text:
Leading edge or trailing edge stall?
As I wrote last week, I'm doing a Benedek 10355 airfoil performance report.
At this point of the work I am analyzing the stall of the profile at a fixed Reynolds number.
The Benedek 10355 airfoil has a thickness of 10.1%.
Figure 1 shows the polar calculated with Xfoil at Reynolds number = 500,000.

Analyzing this curve, and comparing it with this answer, it seems to me that we are in the presence of a trailing edge stall.
In fact, after the stall angle ($\alpha_{stall} = 11.0°$), the decrease in the lift coefficient is gradual ... Then, however, at about $\alpha = 21°$ there is a sudden change in lift, as if a laminar bubble had "burst".
Figure 2 shows the separated zone for $\alpha = \alpha_{stall} = 11.0°$.

Looking at the pressure coefficient, there appears to be a laminar bubble very close to the leading edge (Figure 3).
As the incidence increases, this bubble moves more and more towards the leading edge and, when $\alpha = 22°$, it "disappears".
This makes me think that maybe at low incidences there is a trailing edge stall and then, going further, a leading edge stall ...
Maybe I'm missing something?
Thanks.

So what you're asking is where flow separation's happening?

@CamilleGoudeseune, yes I am asking where the laminar flow separation is happening and if we can talk about a combined leading edge-trailing edge stall.

The Benedek 10355 clearly stalls with separation starting from the trailing edge.
There is some discussion whether a laminar-to-turbulent transition due to rising pressure will always involve a laminar separation bubble, however short. But that bubble does not constitute a leading edge stall.
XFOIL is not meant for flow simulation deep into the stall region. It does amazingly well to capture the initial stall but cannot possibly model the extended separated flow well into the stall accurately. I am actually surprised that it still produces converging results at such high angles of attack.
In short, I would not expect the XFOIL results above 12° AoA to model accurately what is happening in reality. Especially that sharp drop in lift at around 21° looks very much like the code settling on another solution but not something that can be expected in real life.

thank you for helping me this time too. What do you suggest is best for my report? Stop at 12-13° or show this "anomalous" Xfoil behavior? I have to do a post-stall study, that's why I'm interested in going beyond $\alpha_{stall}$. Thanks again

@leo95nf: I wonder who told you to do a post-stall study with XFOIL. Of course you may describe what you found, but be sure to add a disclaimer that real results may very well be different.

Thank you again!


text:
How do I set up G1000 GFC700 autopilot for takeoff?
I want to be able to engage autopilot right away at a safe altitude, while taking advantage of flight level control at Vy for the climb.
Heading is set for the runway. I’d switch to NAV mode when I was ready to make the turn out.
Altitude is set for cruise altitude.
Does this look like I’m all set to just hit “AP” at a safe altitude, or am I missing anything? Thanks!

Is this for a simulator (what type?) or for an actual aircraft?

This looks like a proper setup for an autopilot controlled climb up to 5000 feet. One way to know for sure is to see what the Flight Director tells you. Does it seem reasonable and correct?

Thanks guys. This is a simulator screenshot of course, but I mainly fly a Diamond DA40 in real life.
Great answer Dean!

This looks like a proper setup for an autopilot controlled climb up to 5000 feet. One way to know for sure is to see what the Flight Director tells you. Does it seem reasonable and correct?
Also, prior to taking the runway, you should test the autopilot disconnects. Some of the more common ones are:

Can you overpower the autopilot manually.
The Control Wheel Steering button
The Autopilot immediate disconnect button on the yoke.
The electric trim buttons on the yoke.
The Take Off Go Around button on the panel.
The AP button itself
The Autopilot circuit breaker (this can only be tested during preflight inspection since completely restarting the G1000 will be necessary).


text:
Is the longitudinal stability achieved the same for all aircrafts or does it differ?
If there is a difference in longitudinal stability how are they achieved?

Most simply, by adjusting the relative position of center of mass and aerodynamic center.  Wikipedia's page is a good starting point, actually.

Does this answer your question? How is an aircraft designed to be longitudinally stable?

Different types of aircraft do have differing requirements for longitudinal stability:
A dogfighter or aerobatic sportster needs to be minimally stable, even a little unstable, for maximum manoeuvrability. On the other hand a primary trainer must be stable, so the novice pilot finds it easy to fly.
Whether the plane is fly-by-wire also makes a difference; relaxing the stability reduces drag and therefore improves fuel efficiency, but it needs a computer to add artificial stability so the pilot does not get tired wrestling with the controls for hours on end.
Technically, longitudinal stability is determined by the position of the centre of gravity relative to the aerodynamic centre. If the CG is in front then the plane is stable, if behind then unstable. The designer has to be careful to attach the wings in the right place for the particular characteristics required.
Some planes can vary significantly in flight. The de Havilland DH.88 Comet racer had a big fuel tank in the nose. When full of fuel it was just stable, but as it emptied during a long flight the CG moved progressively aft and it became more and more unstable (and the pilot became more and more exhausted!).


text:
How is a transonic aircraft longitudinal stability different than a subsonic and supersonic aircraft?
/I would like to know-how is the longitudinal stability is achieved between different aircrafts

Why do you keep asking the same question?

Does this answer your question? How is an aircraft designed to be longitudinally stable?

At this level, there is nothing special about transonic stability.
Generally, airplanes become significantly more stable at supersonic speeds (due to aerodynamic center moving backward). At the same time, elevator efficiency drops. Transonic is simply the transition between the less stable subsonic regime and more stable supersonic regime.
Now, for an engineer as well as a pilot, transition can be a challenge, especially if it's quick. For example, most supersonic aircraft have variable stick-to-elevator ratio, and transition happens at transonic speeds. Apart from making sure that the aircraft is controllable throughout the transonic range, you need to consider what happens if the system fails at any stage.
As another example, modern fighters are often made longitudinally unstable. But this is only true at subsonic speed. They are all stable at supersonic regimes. Now you have transition between even more different states: unstable to stable (and back). Flight control systems must cope with that.
Thankfully, the transition is not too abrupt, especially for swept wings. (This, controllability, is in fact one of the main reasons to use swept wings. It's not just about drag).

It is not different.
Aircraft is longitudinally stable if increase in angle of attack causes larger increase in coefficient of lift of the aft horizontal surface than in the forward horizontal surface, which happens if the forward surface flies at higher angle of attack than the aft.
For most aircraft the forward surface is the wing and the aft surface is the horizontal stabilizer, but in canard layout the canard is the horizontal surface and is in the front. For tailless aircraft it is the forward and aft part of the same wing with carefully set sweep and twist.
The angle of attack of the horizontal stabilizer is adjusted to control the aircraft—note that changing camber by deflecting the flap (elevator) at the trailing edge is equivalent to changing the zero-lift line and therefore principally similar. In most aircraft, the horizontal stabilizer actually flies at negative angle of attack to provide sufficient range for adjustment while staying in the range for static stability, but the condition is that it is less than for the forward wing, not negative.
The only difference for transsonic and supersonic aircraft is that when using supersonic wing, the centre of pressure shifts from about quarter-chord to mid-chord when the flow speed over the wing exceeds speed of sound. This causes strong nose-down moment (known as “Mach tuck”) and requires significant change in the incidence of the horizontal stabilizer (decrease for conventional tail, increase for canard). It also creates irregularity in the control behaviour—you push down/trim nose down to accelerate, but then you need to pull up/trim nose up when you reach the critical speed.
However this depends on the flow speed perpendicular to the wing, so sweeping the wing increases the speed at which the Mach tuck occurs. And transsonic, and even most supersonic aircraft have the wings swept so that it does not happen in their design speed range.
So for modern aircraft you don't need to bother and there is indeed no difference. It was significant for older aircraft with straight wings when they tried to break speed limits, X-1, the first aircraft to “break the sound barrier” that still had straight wings, and some fighters with supersonic wings like the F-104.

Something is wrong in "if increase in angle of attack causes larger increase in coefficient of lift of the forward horizontal surface..." It should be the other way round, and in relative terms. And it doesn't explain tailless aircraft. It would be better to say more generally that a pitch moment arises that counteracts the AoA change.

@Zeus, thanks, it was indeed wrong way around.


text:
How to properly compute the stable vertical path load factor
I've been told by someone else that in a stable flight when the vertical path of the aircraft is maintained, the Nz value is equal to cos(pitch)/cos(bank), but I can't seem to figure out how that is derived, the only place on the whole internet where I was able to find that info was here: Inside another stack exchange post, however, it didn't explain how was the formula derived. I do understand that inside a banked turn with 0 disturbance to the vertical path the aircraft's load factor is defined by 1/cos(bank) as explained in the textbook I read: Introduction to Aerospace Engineering with a Flight Test Perspective.
But this doesn't explain the cos(pitch) part of the formula, and nowhere was I able to find anything related to it on the internet. Can someone please explain this to me? Thanks!

Intuition: load factor in a straight unaccelerated climb is cos(pitch). So in turning unaccelerated flight we simply multiply the load factor for climb and for turning. However I'm not sure how to prove this intuition more rigourously.

@Sanchises "chandelle"

@Robert A chandelle comes with a decrease in airspeed so it's not unaccelerated.

Let's look at the extreme case of a 90° pitch attitude. The airplane climbs vertically. Thankfully, in case of an F-16 without loads this is possible (left side of sketch below).

The vertical load factor is defined in the body axis system (blue arrows): $n_z$ is the multiple of gravitational acceleration that presses the pilot into his seat. With $\Theta$ = 90° no gravitational acceleration along the body z axis is left (red arrow), all acts along the body x axis. So here $n_z$ is zero and $n_x$ is -1. Gravitational acceleration presses the pilot into his backrest, but his behind is unloaded.
Since the cosine is 1 for 0° and 0 for 90°, it describes the variation in $n_z$ over pitch. For 180° the cosine becomes -1 and the pilot will hang in his shoulder straps. And so on. The right side of the sketch shows an intermediate case and it should be immediately clear that the component of g in z direction is sin(45°) = 1/√2.
Combine this with the change in $n_z$ for a turn and the formulas
$$n_x = -\frac{sin\Theta}{cos\phi}$$
$$n_z = \frac{cos\Theta}{cos\phi}$$
should make sense. Since the x axis is defined as positive in flight direction, the x component of the load factor is negative with positive pitch attitude, hence the sign of the equation for $n_x$.

Is there a convention that (large) Nz corresponds to an aircraft referenced z axis?   (Since large L is). The diagram seems to show "z" as an earth referenced vector.

@RobertDiGiovanni Z is moving with the aircraft - how can it be earth referenced?

The answer would imply a flat earth. Thanks to you and Sanchises, we now have a combined formula for lift requirement plane referenced.  I am finding it much easier to decompose the weight vector, and leave the (aircraft generated) lift, thrust, and drag vectors whole.

my friend has confused neutral vpath G load with a V-n diagram G load and thinks that the G load increases with airspeed, can you please so the difference between the two. Also, the answer above is pretty good I just didn't quite get why Nx is  = -sin(theta)/cos(phi), anyway thanks for the answer.

@JonathanOrr The possible g load does indeed increase with airspeed, but then you have a nonzero pitch rate which needs to be added to the equations above.

In a stable level turn vertical component "z" will equal the weight of the plane (in order to maintain altitude).  Assuming no change in velocity, the AOA must be increased by AOA/cosine bank angle (as long as the plane does not exceed stall AOA).
G load in a turn = 1/cosine bank angle.  Increase in coefficient of lift is generally linear to increase in AOA.
A refinement to the reasoning might be to consider the increase in AOA from level flight AOA. For example:  if 1/cosine bank angle = 1.5 and my AOA is for level flight is 6 degrees, my AOA in the turn is 9 degrees.  If the plane stalls at 12 degrees, this turn is safe.
("AOA level flight" would be the difference of AOA zero lift and AOA required for level flight.  Some wings have a negative zero lift AOA).
To summarize mathematically for level flight:

AOA turn/AOA level flight = 1/cosine bank angle

But...

For a stable flight when the vertical path is maintained

This implies a climb, level, or descending flight! Here, for all three cases, lift = weight x cosine pitch (to the horizon).
So, we may include "cosine pitch (to the horizon)" in our lift and turn formula as Nz = cosine pitch/cosine bank angle.
Amazingly, because lift is less than weight in a climb, the stall speed is lower.  The F-16 is perfect for this concept, as it has the excess thrust to make it happen.

-1 answer does not contain a derivation for the load factor n=cos(pitch)/cos(phi) as per the question.

please see edits


text:
Can I use my FAA Private Pilot (Foreign Based) without a valid EASA SEPL to fly in USA?
Can I fly in the US using my FAA Private Pilot (Foreign Based) based on my EASA PPL without having renewed SEP(L) - it's EASA's class rating for single piston engine airplanes. By flying I mean using my FAA ASEL rating.

This question has already been answered.  The answer is NO.

The answer is YES - read carefully proposed post and letters mentioned in the answer - "The FAA views aircraft ratings as privileges for operating certiain aircraft." - from Grossman letter


text:
Pilatus PC-12 vertical tail size
Does anyone know what the dimensions / surface area of the vertical stabilizer is on that plane? Additionally, does anyone know the approximate tail volume and its coefficient?
Could not find it anywhere on the internet.
I do not need exact figures (although it would be better), educated guesses or approximations are enough.

You could get a rough estimate from a scale drawing.

Or by measuring a picture in profile. Finding the overall length should be easy enough, so from that you'll have the scale of the picture you find.


text:
Size of speed brakes for glider
I have a concept glider and I am trying to calculate the required speed brake area. The clean glide ratio is currently 30, and with speedbrakes, I am aiming at reducing it to something around 7.
I know that the Cd of the speedbrakes is around 1.5.
As for other specs, the wing area is 12 square meters, the AR is 18 and the glider weighs 400 kilos clean.
How can I calculate it?

Take the Cd  of the glider clean, then work the speed brake drag based on its area.  Glide ratio  × weight gives you your needed total drag.

Most modern gliders, as a safety feature, have "speed limiting brakes", which means if you pull them all the way out the glider can't exceed Vne in any attitude. Your target drag should be sufficient to keep the glider from exceeding Vne in a vertical dive.


text:
What exactly does "efficiency" mean in XFLR5 wing analysis?
Can somebody please tell me what it is / how is it calculated?

I initially thought it might be something with the Cl/Cd but I have another design with close to the same Cl/Cd, but with more than double the efficiency (more than 1.0)

Try various angles of attack up to stall.  Your data shows AoA of 0 degrees.  Go higher AoA.  CL/CD may improve.

@RobertDiGiovanni it does, but that does not answer my question

Could it be Oswald efficiency?

@RobertDiGiovanni: I also think it is the Oswald coefficient aka efficiency factor. @ Vincent can you plot the spanwise lift distribution?


text:
How to calculate boundary layer thickness?
So I have a flat plate with a surface of 7sqm and an aspect ratio of 12 with air blowing straight at it (like a wing) at a speed of 50km/h. I would like to manually calculate the thickness of the laminar boundary layer at the trailing edge of the plank.
How do I do that?

Look up the Blasius solution for the laminar boundary layer.
There are three measures of boundary layer thickness -- the plain thickness, the momentum thickness, and the displacement thickness.  There are equations for all three available.

thank you, I used the 99% method I found on this page: https://en.wikipedia.org/wiki/Boundary_layer_thickness#The_bounded_boundary_layer_description


text:
how does the lilium jet counter asymmetric thrust?
So Lilium states that its jet is a tailless design. Now, you can see that it has engines on both sides of the centerline.
Lets imagine that all of the engines on one side fail, how does it counter that asymmetric thrust?
We could speculate that their jet actually has a tail, if you look at the landing gear (It is kinda shaped like a vertical stabilizer - ignoring the fact that I cannot see a rudder there), but even then, it does not seem big enough to counter that thrust.
The winglets could be used as one, but again, I do not see any rudders there.
So how does it counter asymmetric thrust?

Has this aircraft actually flown? If not arn't we asking for a theoretical answer to a theoretical question?

@Jamiec: It has, actually. It is currently in prototype flight trials.

My assumption is: it doesn't. Most multi-engine airplanes are designed to survive the failure of one engine, maybe two. The situation you refer to requires 18 engine failures.

It did test-fly indeed. According to lilium's web page the systems are built redundant so a case like that should never happen - obviously unless something definitely worse has happened but then there would be another issue to be managed. Anyway in that case the full aircraft parachute should allow a safe landing.

@sophit it sounds like you can actually write a referenced answer then. (referring to  "According to their web page")

@Jamiec: unfortunately there's nothing more than that. To correctly answer, one should know the electrical connections among engines and batteries, at least schematically, but I don't think that this information is made publicly available. Therefore answering this specific question would be pure speculation at risk of deletion.

This question is a bit like "what if all 18 wheels fall off a tractor trailer at once".  Lilium is electric, which means this type of failure is highly unlikely.  If they are in hover, it would cartwheel.  In flight, the other side could be shut down for an emergency glide.  Note the canard also has 2 banks of engines.  Some directional control may be possible in that case.

@sophit "answering this specific question would be pure speculation at risk of deletion." -- if this is the case, why haven't you voted to close as opinion based?  I see it as a reasonable question, answerable on the basis of the capabilities of what we do know about the aircraft (hence my answer below).

@sophit you should still be able to see the comments on your deleted answer, if you have enough rep.  Looks like you asked for deletion after a downvote.  If you can't see it now, however, it's hard to ask for undeletion.

@ZeissIkon: oh well, a couple of hours ago this very question was indeed closed, don't know why it was reopened. And I also wrote an answer that was deleted as well... Apparently I'm not that good in answering since your is still here an gets upvoted

@ZeissIkon: I anyway agree with the comment of RobertDiGiovanni

@ZeissIkon: and I agree with you that this is a reasonable question, that's why I upvoted it and gave an answer

Since Lillium's "engines" are electric ducted fans under computer control (for thrust control and balancing for yaw, and pitch when in hover) anyway, it would be trivial for the computer to match thrust reduction on the opposite wing to any reasonable level of failures.
Say two fans on the left wing ingest parts of a bird big enough to destroy the fan disk itself -- the computer can detect the loss of load on those two motors, shut them down, and either shut down corresponding motors on the right wing, or reduce thrust on all the right wing fans to compensate.  As noted in comments on the question, the two fan arrays on the canard can also be used in compensating yaw, with less authority but also with less effect on overall thrust.
Good design would suggest the aircraft be designed to hover with at least two (or more) fans out on each wing or canard, as well, just because one of the most likely failures (FOD/bird strike) has a high chance of destroying more than one fan simultaneously due to their close-packed installation.

so they are basically betting that a situation of one side failing completely never occurs

@VincentCerowski Just as all those piloted quadcopters and even the V-22 Osprey do (yes, it's got crossover shafts and gearing, but if the gearbox in one engine fails in hover, they're done).

@VincentCerowski: Just like conventional aircraft are betting that all 4 engines, all 3 hydraulic systems, all 3 electrical systems, and all 5 flight computers failing while all three pilots simultaneously die of a heart attack never occurs.


text:
XFLR5 vs AVL vs DATCOM vs OpenVSP
I am working on a model airplane and am coputing some of the data using software.
I have already computed the general stuff using XFLR5 and AVL. My question is, is there any point for me to do the same in DATCOM and OpenVSP? To rephrase my question: Will they offer me more results / results for other different conditions? Or will I get pretty much the same stuff I got in the former two programs?
Or maybe this question will be better: What is difference between those programs? What more do they offer over eachother?
Thanks and have a great day!

It depends on what you want from the calculations.  First, I'll assume that you are mainly interested in calculating the pitch stability of the aircraft to help you determine where to locate the CG.
The tools you've used are fine and you won't gain much from the other tools.
AVL, XFLR5, and VSPAERO use fundamentally similar (though with differences) techniques -- i.e. they basically amount to a thin-surface vortex lattice potential flow code.
DATCOM instead uses a component-wise buildup and other table lookups from classical computations (some may be similar to a VLM) and empirical fits (from test data) to calculate stability derivatives.
XFLR5 is essentially AVL with a boundary layer model added on (like XFoil).
VSPAERO also has a thick-surface mode, but it won't make a big difference for the computations you're after.
VSPAERO has the ability to include an actuator disk or unsteady rotating blade propeller model.  If you needed to capture propeller effects, that would be a strong advantage.
VSPAERO has a linearized supersonic mode.  VSPAERO has some very limited viscous models.  VSPAERO has unsteady modes and other bells and whistles.
DATCOM has results for transonic, supersonic, and hypersonic flows -- it also includes viscous effects.
The most important difference between the tools is how easy it is to build the input model.  OpenVSP (I'm admittedly biased) is far and away the easiest here.  You can go from nothing to a model and results very quickly.
XFLR5 and AVL have very similar approaches to build the model.
DATCOM is a very different world.  If you're using a user-friendly front-end software to DATCOM, it may not be too bad, but if you're classically using the paper copy of DATCOM, it is a laborious process to get a full set of results.
If your aircraft is conventional, all of these are easier.  If for example, you're designing a flying wing, then you'll need to be more careful to make sure the lateral-directional effects are captured.
The best place to ask OpenVSP questions is on the OpenVSP Google Group.

XFLR5 is not recommended when u are using it for high subsonic conditions.
The software did not iterate to match the mach number.
It is only valid for incompressible flow.


text:
How can I calculate the TOTAL drag of the ENTIRE airplane?
I have an aircraft similar to a Cessna 172 and am using a virtual wind tunnel software (flowdesign) to get a drag estimate. I import the airplanes 3D model and I get around 2240 lbs of drag with a drag coefficient of 0.33 at a speed of 253fps, which I think is way too much to be trusted.
I then tried to calculate it using an equation (although the equation is supposedly only for wing drag): D=q∗S(Cd0+CL^2/(π∗AR))
Where q is dynamic pressure, S is the wing surface area (208sqft in my case) and AR is the wing aspect ratio (6). I used a D/q value of 7 (to be conservative) and a Cd0 of 0.033. With that, I got a result of 560lbs of drag (remember, this is supposed to be just for the wing) at a speed of 253fps (which is the speed at which the best L/D ratio is supposed to be according to the equation - and is the same speed I used in the virtual wind tunnel)
With that result, I then went to calculate how much thrust I would need for a climb rate of 1250fpm. The equation is: T=W*(Vv/V+D/L)
Where W is the weight of the airplane, Vv is the targeted vertical speed (1250fpm), V and D are the speed and drag for the best L/D ratio (so 560lbs and 253fps), L is probably the lift and I used the same value as the weight there. With that, I got the result of 923lbs of thrust needed. That is around 723 horsepower.
Now if I go look up the horsepower of a Cessna 172, google says it is 180. The climb rate is 720fpm, which is half of what I targeted for, but the difference is still too big for my calculations to be right. The wing area is 175sqft, only about 30sqft less than I have.
I then tried to calculate the drag with another equation, in which I additioned the induced and parasitic drag. The equation for the induced drag is: Di=Cdi*S*q
Where Cdi=Cl^2/(π∗AR) where Cl=W/(q*S) and I got a result of about 118lbs of induced drag. For parasitic drag, I used Dp=D/q*q and got around 284lbs of parasitic drag. The total drag is 403lbs, which I believe is still way too much to be true.
I then tried using the wright equation, which is: D=Cd*q*S, and I got a result of 3160 lbs of drag using a Cd of 0.033 (now I know I probably should have used something closer to what I got from the virtual wind tunnel, but the result would have been even greater) and a total area of 3160sqft (that is the total wetted area of the whole airplane).
All of the above results could be countered with enough horsepower (and that is somewhat realistically doable), but when I compare the horsepower required, the climb rate and the wing area to other already existing planes, my calculations seem off.
So what is a simple way of calculating the total drag of the entire airplane accurately?
Sorry for the wall of text.
P.S.: if anyone is interested, I am using these videos (or series of videos) to conceptualize an airplane design:

 ,

So, you are basically calculating the drag with 3 equations which are the same and you get 3 different results. I suggest you to stick with a consistent system of units (best SI) and recheck the calculation. And get some standard book about general airplane design (Raymer for example) instead of online tutorials: they are really easy to read and full of (correct) numbers for any aerodynamic coefficient

Have also a look to this questions:
https://aviation.stackexchange.com/questions/43410/how-is-the-zero-lift-drag-coefficient-calculated
https://aviation.stackexchange.com/questions/56629/how-to-calculate-the-induced-drag-coefficient?rq=1

Does this answer your question? How is the zero-lift drag coefficient calculated?

EDIT
Doing the math at a speed of 149kts (as suggested by @RobertDiGiovanni) and a weight halfway between EW and MTOW, gives a total drag of some 308lbs which should be a  reasonable value.

Try with:
$C_{D_{total}}=0.02+0.05C^2_L$
this should give you a reasonable target for your simulations.

How can I calculate the TOTAL drag of the ENTIRE airplane?

Total drag polar for an aircraft can be calculated as the sum of 1. the polar of the wing plus 2. the drag of all the other components:

$C_{D_{wing}}=C_{D_{min}}+ k(C_L-C_{L@C_{D_{min}}})^2$
$C_{D_{rest}}= C_{D_{fuselage}}+C_{D_{landing gear}}+C_{D_{nacelle}}+…$

where:

$C_{D_{min}}$, $k$ and $C_{L@C_{D_{min}}}$ depend on the particular airfoil(s) used for the wing and its geometry (sweep angle, aspect ratio, twist, ...).
each term of $C_{D_{rest}}$ is normally estimated using known standard historical values, scaled via the basic dimension of each component.

Each and every term depends also on the speed i.e. on Reynolds and Mach numbers. Drag terms of 2. may also depend on lift (landing gear for example).
The sum of 1. and 2. gives the following general form for the drag coefficient:
$C_{D_{total}}=a+bC_L+cC^2_L$
Normally the second term $b$ is neglected and the total drag coefficient assumes the well known form:
$C_{D_{total}}=C_{D_0}+KC^2_L$
For a single-engine light-aircraft with symmetrical airfoil and flying at subsonic speed, plausible rough values for $C_{D_0}$ and $K$ based on historical values are:
$C_{D_{total}}=0.02+0.05C^2_L$
Anyway a better approximation (within a ±10% error) can be achieved only evaluating each term of 1. and 2. I suppose that doing the math here for one particular case is beyond the scope of this Stackexchange, also because it would take some time and space. The values to be used and the general theory behind it can be found in any standard book of general airplane design, like the one by D. P.  Raymer, Dr. J. Roskam, E. Torenbeek or L. Nicolai. All this books are really easy to read, fluent and with almost no equation.

Running the calcs at high cruise finds us squaring a Cl value of less than one.  Why don't you show one calc at 149 knots.  It'd be worth an upvote perhaps.

@RobertDiGiovanni: I don't believe in doing the homework of somebody else (not referring to you), I prefer to give the tools and show the path ;-)  Anyway I'm going to do it, I just need some time

Thanks for the answer. Also, you were right in your comment under my question. My dumbass forgot to convert from knots to fps (should have been 150 fps, not knots). I have redone my calculations and now the result seem more reasonable. I also want to thank @RobertDiGiovanni for his answer, having something to compare my results to is very useful as a first rough "reality" check.

:-D glad I could help  And seriously consider getting one of those books, older version can be found almost for free

Drag of the $entire$ airplane is readily available by gliding, or monitoring fuel consumption.
Checking fuel consumption is a bit more complicated because what you really need is thrust.

in steady state flight, thrust = drag

But just to get "in the ball park", a 172 weighing 2400 lbs gliding 8:1 generates 300 lbs of drag at Vbg.  That is about how much thrust you need for level flight, with around 300 lbs more available to climb.
Of course these numbers will vary depending on airspeed and configuration but we can see even 923 lbs of thrust is a bit high.
253 feet per second comes out to 149 knots$^1$!  Best L/D for a 172 is around 65 knots, or around 110 fps.  Try that number and see what you get.
$^1$ keep in mind "cruise" numbers in POH are true airspeed, where higher altitude will help.  403 pounds of total drag here seems to be closest to reality, especially considering decrease in fixed prop thrust at that airspeed.


text:
Does leading-edge droop increase or decrease lift, other things being equal?
Consider a conventional subsonic airfoil in level flight. Typically it is cambered and with a small positive incidence, both of which features contribute to lift.
Now droop the leading edge, as with say a plain leading-edge flap. This reduces the incidence, possibly even taking it negative, but increases the camber. Will the net lift increase or decrease, and why? Can the answer vary according to other flight conditions?